Problem: Let $x,$ $y,$ and $z$ be positive real numbers such that $x + y + z = 1.$  Find the minimum value of
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z}.\]
Explanation: By AM-HM,
\[\frac{x + y + z}{3} \ge \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}.\]Hence,
\[\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \ge \frac{9}{x + y + z} = 9.\]Equality occurs when $x = y = z = \frac{1}{3},$ so the minimum value is $\boxed{9}.$